Separation Processes:

MembranesChE 4M3

© Kevin Dunn, 2014

kevin.dunn@mcmaster.ca

http://learnche.mcmaster.ca/4M3

Overall revision number: 308 (October 2014)

1

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2

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3

Administrative issues

Project topics are posted. There are 3 main topics:

1. commercial, large-scale, production of membranesI how are membranes produced?I modifications to operate at higher temperature;I modifications made to the surface properties to improve its

performance and reduce fouling.

2. separative reactorsI reactor + separation combined together and integrated;I operation of the unit;I use of separative reactor technology at a large scale;I economic impacts.

3. bioseparations using chromatography and/or ion-exchangeI what are the challenges in operating these units?I how are the resins and adsorbents typically selected?I some terms: capacity; selectivity; specificity; regeneration;

loading and unloading.

Due by 19 November 2014, at 9:30

4

Membranes

[Flickr: 21182585@N07/2057883807] [Flickr: 21182585@N07/3574729377]

5

Quick activity

On a sheet of paper write

I bullet points and/or

I draw a diagram and/or

I describe

“what you know about membranes”

6

References

I Perry’s Chemical Engineers’ Handbook, 8th edition, chapter20.

I Wankat, “Separation Process Engineering”, 2nd edition,chapter 16.

I Schweitzer, “Handbook of Separation Techniques forChemical Engineers”, chapter 2.1.

I Seader, Henley and Roper, “Separation Process Principles”,3rd edition, chapter 14.

I Richardson and Harker, “Chemical Engineering, Volume 2”,5th edition, chapter 8.

I Geankoplis, “Transport Processes and Separation ProcessPrinciples”, 4th edition, chapter 7 (theory) and chapter 13.

I Ghosh, “Principles of Bioseparation Engineering”, chapter 11.

I Uhlmann’s Encyclopedia, “Membrane Separation Processes, 1.Principles”, DOI:10.1002/14356007.a16 187.pub3

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Why use membranes?Some really difficult separations:

I finely dispersed solids; density close to liquid phase; gelatinousparticles

I dissolved salts must be removed

I non-volatile organics (e.g. humic substances)

I biological materials: sensitive to the environmentI biological materials: aseptic operation is required

I cannot centrifugeI cannot sediment

It is usually worth asking:

How does nature separate?

I energy efficient

I effective

I maybe slow?

8

Why use membranes?Relatively new separation step (“new” meaning since 1960 to 1980s)

I often saves energy costs over alternative separationsI ambient temperature operation

I often easier to operate and controlI more compactI lower capital cost than alternatives

Modules:

I feed stream split into parallel units

I easier to maintain and replace parts

I can be expanded as needs grow

[Henk Koops’ slides, GE Water and Process Technologies] 9

Challenges in membrane design

Challenges that still remain:

I withstanding high pressure differences but still have a thinmembrane

I dealing with fouling and cleaning

I increasing selectivity (separation factor) for specificapplication areas

I uniformity of pore sizes

I temperature stability (e.g. steam sterilization)

Membrane manufacture is a complex area: very fruitful area forpolymer engineers

10

Market size

[Perry’s: chapter 20, 8ed]

11

Let’s formalize some terminology

[Seader, Henley and Roper, p 501]

12

More terminology

semipermeable: partially permeable, e.g. your skin allows certainsize particles in, but not others

mass separating agent: the membrane itself

energy separating agent: the applied pressure (pressure drop)

porosity =area of open pores

total surface area

13

What is flux?

The (volumetric) or (molar) or (mass) flow per unit time for 1 unit of area

I J = flux =transfer rate

transfer areaI e.g. 42 mol.s−1.m−2

I never simplify the units: write 13(m3.s−1

).m−2

I you may, and probably should, omit thebrackets: 13 m3.s−1.m−2

I do not write 13 m.s−1

General principle

For a given unit area, we want the highest flux possible (at thelowest possible cost)

14

Membrane classification

[Richardson and Harker, p 438]

15

Transport through a membrane

Why study theoretical models?

All forms of membrane applications rely to some extent on thesame equation structure. The details will change.

Will allow us to:

I troubleshoot problems with the process

I predict expected impact of improvements/changes to theprocess

I used for crudely sizing the unit (order of magnitude estimates)

16

Examples you will be able to solve

1. how long should we operate unit at constant ∆P to achievedesired separation?

2. what is the mass transfer coefficient through the labmembrane?

3. what pressure drop (and therefore pump size) do I expect?

4. how many cassettes (area) does this application require?

17

The general equation

transfer rate

transfer area= flux =

(permeability)(driving force)

thickness=

driving force

resistance

Symbolically:

ρfQp

A=ρfA·dVdt

= J =(permeability)(driving force)

L=

driving force

R

I permeance =permeability

L=

1

resistance=

1

R= “mass transfer coeff”

I permeance: easier to measure

I permeance units: depend on choice of (driving force) and J

I resistance = f (thickness L, viscosity, porosity, pore size)

I we will specifically define resistance in each case

18

MicrofiltrationI 0.1µm to 10µm pores: sieving

mechanism

I conventional filters: not effectivebelow ∼ 5 µm

I microfiltration membranes: generallysymmetric pores

I polysulfone membrane

I (surface) porosity as high as ε = 0.8

I driving force = ∆P: 100 to 500 kPa

I high fluxes at low TMP(trans-membrane pressure)

I application areas:I yeast cells harvestingI wine/beer/juice clarificationI bacteria and virus removalI air filtrationI cytology: concentrate up cells

[Henk Koops’ slides]

19

General modelling equation applied

ρfA· dVdt

= Flux = J =∆P

µ (Rm`M + RcLc)

ρfA· dVdt = Flux = J =

∆P

µ (R ′m + R ′c)

J [kg.s−1.m−2] permeate fluxµ [kg.m−1.s−1] permeate viscosity

∆P [Pa] = [kg.m−1.s−2] TMP varies for different applicationsRm [m.kg−1] resistance through membrane (small)Rc [m.kg−1] resistance through cake (large)`m [m] membrane thicknessLc [m] effective cake thicknessρf [kg.m−3] fluid densityR ′c [m2.kg−1] “cake resistance”

[Illustration from Richardson and Harker, Ch8] 20

Flow patterns for microfiltrationDead-end flow

I only for very lowconcentration feeds

I else becomes rapidlyclogged

I air filtration and virusremoval applications

J =∆P

µ (R ′m + RcLc)

Cross-flow (TFF)

I TFF = tangential flow filtrationI main purpose?

I microfiltration: tends to havecake build up

I induces shearing to erode cakeI mobile slurry for downstreamI reduces cake resistance, R ′

c

I ∆P =Pin + Pout

2− PP

21

Dead-end flow vs cross-flow geometries

Dead-end flow

I cake thicknessincreases with time:Lc(t)

I implies cakeresistance changeswith time: R ′c(t)

I so for a constant ∆P,implies J(t) falls off

J =∆P

µ (R ′m + RcLc)

Cross-flow (TFF)

I fluid velocity: 1 to 8 m.s−1

tangentially

I keeps mass transfer resistance low

I for a given ∆P: TFF allows us toobtain higher fluxes than dead-end(usually ∆P is 100 to 500 kPa)

I cannot take lab test results with afilter cloth dead-end and apply it tocross-flow situation

22

Cross-flow flowsheet

[Illustration from Richardson and Harker, Ch8]

How to pressurize the unit?1. Supply feed at pressure; valve at retentate to adjust/control ∆P2. Draw a vacuum at permeate and pull material through membrane

Question: why recycle the retentate stream?23

Dealing with fouling

a. slow cross-flow velocityb. high cross-flow velocityc. high cross-flow with regular backwashing

[Richardson and Harker, Ch8] 24

Factors to improve flux

I increase pressure difference

I regular backflush

I choose alternative membrane structure

I feed concentration kept low

I shear rate (velocity in cross-flow): reduces R ′c = RcLcI increase temperature of feed

I nature of the solids deposited: affects resistance Rc

25

Question

A microfiltration membrane operating with pure feed of waterproduces a flux of 0.06 kg.s−1.m−2 when operated with a TMP of150 kPa.

1. What is the resistance due to the membrane? Specify theunits.

2. If operated with a protein-water mixture at a 200 kPa pressuredifference, a flux of 0.0216 kg.s−1.m−2 is measured at steadystate. What is the resistance due to cake build-up? Specifythe units.

3. Next, estimate the pressure drop required to achieve a flux of0.035 kg.s−1.m−2 [Ans ∼ 325 kPa].

To consider: are we likely to achieve fluxes of 0.1 kg.s−1.m−2

with this membrane? If not, how could we?

26

A very crude estimate of the membrane resistance, R ′mI Assume the pores are cylinders of diameter D with length `MI The velocity in this tube for pure solvent is

v =(D2)(∆P)

32µ`MHagen-Poiseulle

I Consider a 1 m2 surface of tube pore openings, where ε is thefraction of pore openings of diameter D

I The total flux of solvent through all pores is

J = v ρf ε =(D2)(∆P)

32µ`Mρf ε =

∆P

µ

(32 `MρfD2 ε

) =∆P

µR ′m

I Note: true pores are of different sizes, they are not straightthrough the membrane; they bend and twist

I Note: this R ′m will always be too low

27

Estimating the cake resistance, Rc : for dead-end filtration

Exactly the same approach as we saw in the filtration section:

1. Measure V (volume of permeate) against time t at a constantand known ∆P

2. The slope is related to α, the specific cake resistance [m.kg−1](we defined this in the filtration section)

3. R ′c(t) =αCS

ρf AV (t) will increase over time

28

Ultrafiltration (UF)I 5 nm to 100 nm (0.1 µm) particles are

retainedI 1 to 1000 kDa particles are retained

(move to using molecular weight)I 1 dalton = 1 atomic mass unitI 1 kilodalton = 1000 dalton =

1000 g/molI particles with lower molecular weight,

e.g. most solvents, pass through

I pore sizes: 1 to 20nm

I typical fluxes:Jv = 0.01 to 0.5 m3.m−2.hr−1

Jv = 10 to 500 L.m−2.hr−1 (LMH)

I asymmetric structure

I almost always operated in TFF(cross-flow filtration)

[Perry’s 8ed; Ch20.4] 29

Ultrafiltration applications

UF: loosely considered: “cross-flow filtration at molecular level”

I Recovery of proteins and high molecular weight materials(solute)

I Permanent emulsions: e.g. oil phase will not pass

I Fine colloidal particles: e.g. paint/dyes

I Large molecules of interest might remain in retentate;permeate discarded

I e.g. albumin (egg white) concentrationI e.g whey processing (liquid “waste” after cheese-making):

I centrifuge −→ UF −→ reverse osmosis (RO)I valuable proteins retained by UFI permeate sent to RO to concentrate smaller molecule sugars

and saltsI that RO concentrated permeate: used for ethanol and lactic

acid production

30

More terminology

I permeate: the material passing through the membrane fromfeed to outlet side

I retentate: the material retained on the feed-side of themembrane

I solute: most often retained on the inside (feed side) of themembrane and deposited on the membrane wall

I solvent: the liquid phase that carries the solute

I gel effect: buildup of the solute on the membrane wall to forma high concentration gradient “gel”

31

Ultrafiltration (UF)I driving force = ∆P of 0.1 to 1.0 MPaI “tight”, low-permeability side faces the TFF to retain particlesI this skin layer is about 10µm thick; provides selectivity, SI open, high-permeability side mainly for mechanical strength

[Richardson and Harker, Ch8]

R = 1− Cpermeate

Cfeed

R = 1− Cp

Cf= 1−S

MWCO: molecularweight whereR = 0.9

i.e. 10% of thatmolecular weightpasses through tothe permeate

32

Transport phenomena in UFI solute (i.e. particles) carried towards membrane by solvent

I J =∆P

Rm + Rcp

I Rm = membrane resistance [m.s−1 if J is mass flux]I Rcp = resistance due to “concentration polarization”I Rcp effectively is the resistance due to solute boundary layerI Mass concentration Cb (in bulk), steadily increasing to Cw

(wall)I Units of C are kg solute per m3 solvent

33

Transport phenomena in UF

I Solute flux towards membrane:J

ρf· C = JvC

I Solute flux out of membrane (leakage): JvCpermeate ≈ 0

Net transport of solute = Jv (C − Cp)

Jv

[m3 solvent

m2.s

]volumetric flux of the permeate

C

[kg solute

m3 solvent

]solute mass concentration at a given point

JvC

[kg solute

m2.s

]solute mass flux towards membrane

Cp ≈ 0

[kg solute

m3 solvent

]solute mass concentration in permeate

34

Diffusion termI Solute diffusion away from membrane = Jdiffusion

Jdiffusion = −DABdC

dy

DAB

[m3 solvent

m.s

]=[m2.s−1

]diffusion of solute in solvent

Jdiffusion

[kg solute

m2.s

]solute mass flux into bulk

I See animation on Wikipedia35

Transport at steady stateAt steady state: transfer towards membrane equals diffusion back

Jv (C − Cp) = −DABdC

dy

− JvDAB

∫ Lc

0dy =

∫ Cb

Cw

dC

C − Cp

ln

(Cw − Cp

Cb − Cp

)=

Jv LcDAB

=Jvhw

36

UF: mass-transfer key pointsAssuming Cp ≈ 0 (i.e. R = 1)

JvLcDAB

=Jvhw

= ln

(Cw

Cb

)where hw is a mass-transfer coefficient, with units of m.s−1

I there are correlations forhw = f (velocity, temperature, channel diameter, viscosity)

I when gelling occurs, Cw = Cg at the wallI the effect of increasing ∆P is

I increase in solute flux towards boundary layerI diffusion increases to oppose itI net effect: almost zero (see earlier plot)I experiments mostly agree with this theory

I there is a limiting flux Jlim = f (Cw ,Cb, hw )I at higher feed concentrations, lower fluxes if we are at/near

the gel polarization state (gelling)I typical diffusivities: 1× 10−9 (fast!) to 1× 10−11 m2.s−1

37

Transport phenomena in UFI Experimental evidence agrees well with theory ... to a point.I Increasing ∆P leads to compacting this layer, increasing Cw

I So diminishing returns from increasing ∆PI Also, there is a strong concentration gradientI Diffusion away from membrane due to concentration gradientsI Eventually solute forms a colloidal gel on the membrane, Cg

I Adjusting pressure has little/no effect anymore

[Chemical Engineering Magazine, 8 May 1978] 38

Example question

An ultrafiltration application is required to treat a waste streamthat has 4 kg.m−3 waste in the feed. The desired soluteconcentrate must be 20 kg.m−3.

Pilot plant studies show the flux can be expressed as

Jv = 0.02 ln

(25

Cb

)in units of m3.hour−1.m−2.

1. What is the flux Jv right at the membrane entrance?

2. What is the flux Jv for most of the membrane if we are ableto reach our desired end-point?

3. Interesting: what happens if we require a solute concentrationof 10 kg.m−3?

39

Geometries for ultrafiltrationTubes in a shell

I membrane on a porous support

I cleaned with soft sponge balls

Plate and frame

I batch operation

I All these units bought as complete module from supplierI In fixed sizes; so need to be combined (next section)I Also as cassettes, tubes and flat sheets run in TFF to increase

flux.

[Illustrations from Wankat, 2ed, Ch 16] 40

Geometries for ultrafiltrationSpiral wound

I high surface area per unit volume

I high turbulence, reducing masstransfer resistance

[Illustrations from Wankat, 2ed, Ch 16]

http://www.youtube.com/watch?v=YlMGZWmh Mw: how spiral

membranes are made

Hollow fibre membranes

I largest area to volume ratio

I fibre inside diameter = 500to 1100 µm for UF

I UF: feed inside tube, withthin membrane skin on theinside

41

[Richardson and Harker, Ch8]

42

Sequencing membrane modulesParallel

I most common configuration

I allows increase inthroughput

Series

I used to achieve a desiredseparation factor(concentration)

I high pressure drop acrossseries circuit

I cannot recover pressure(energy separating agent)

[Wankat, Ch16]43

Example of an installation

I Larnaca, Cyprus

I Sea-water reverse osmosis membrane, i.e. desalination

I 21.5 million m3 per year

I parallel and series

[ide-tech.com]

44

Operating UF units

I Continuous operation provides lower-cost operation

I Batch operation: seldom used, except for start up (see next)

I Biologicals: require batch processing to meet regulatoryrequirements

I High solids in feed? Require multiple-pass: simply recycle

45

Recycle operation: “feed plus bleed”

[Modified from Richardson and Harker, Ch8]

I Initially close retentate valve (batch mode operation)I Fluxes slowly reduceI Open retentate valve and operate at steady stateI AIM: achieve a given concentration in the retentate =

Cbulk = Cretentate46

Class example

We need to treat 50 m3.day−1 of waste containing a solute at4.0 kg.m−3. The desired solute concentrate must be 20 kg.m−3.The plant operates 20 hours per day.

At steady-state the feed-plus-bleed circuit operates with a flux

Jv = 0.02 ln

(25

Cb

)in units of m3.hour−1.m−2, where Cb is the bulk concentration,measured in units of kg.m−3.

If each membrane module is 30 m2:

I how many membrane modules are required?

I series or parallel?

47

Multiple units in series

Now consider the previous example.1. Consider 2 membranes in series of 30 m2. What are

concentration and flow values for unit 1 and unit 2?I Note: we do not know the outlet concentration, CR2 this time!

2. Now try find the optimal areas, A1 and A2 for the membranesfor a desired outlet CR2 concentration.

Hint: use a guess-and-check strategy, bearing in mind the modulescan only be purchased in 30m2 units.

48

FoulingI Process feed pretreatment is important.I e.g. in bio area: prefiltration, pasteurisation to destroy

bacteria, or adjust pH to prevent protein precipitationI Backflushing mostly restores permeation rate (opens pores)I Can also use pulsated/oscillating feed flowsI Consider adding tube inserts

I Inject air: sparging with oxygen or nitrogenI Oscillating electrical field works on certain feedsI Chemical cleaning is eventually required [long time], e.g.:

I flush with filtered waterI recirculate/back-flush with a cleaning agent at high

temperatureI rinse to remove the cleaning agentI sterilize by recirculating weak chlorine solution at high tempsI flushing with water to remove sterilizing solution 49

Video

http://www.youtube.com/watch?v=YlMGZWmh Mw: How spiralmembranes are made

50

Reverse osmosis

I One the most requested topics (start of the term!)I One of the largest membrane markets by $ size

1. Dialysis2. Reverse osmosis (water treatment)

I What is osmosis? [Greek = “push”]

I Then we look at reverse osmosis (RO)

I Applications of RO

I Modelling RO

51

Osmosis principle

52

Osmosis principle

53

Reverse osmosis principle

54

(Reverse) Osmosis principle

I Assume solute barely passes through membrane (Cp ≈ 0)

I but solvent passes freely: this is why we call it asemipermeable membrane

I Chemical potential drives pure solvent (water) to dilute thesolute/solvent (mixture).

I This solvent flux continues until equilibrium is reachedI solvent flow to the left equals solvent flow to the rightI results in a pressure difference (head)I called the osmotic pressure = π [Pa]I a thermodynamic property 6= f (membrane)I a thermodynamic property = f (fluid and solute properties),

e.g. temperature, concentration, pressure

55

(Reverse) Osmosis principle

I Osmosis in action:I trees and plants to bring water to

the cells in upper branchesI killing snails by placing salt on

themI why freshwater fish die in salt

water and vice versaI try at home: place peeled potato

in very salty water

I If you exceed osmotic pressure youreverse the solvent flow

I Called “reverse osmosis”

I Net driving force in this illustration=

56

Typical values of osmotic press

For dilute solutions

π ≈ nRT

Vm= cRT

π [atm] osmotic pressuren [mol] mols of ions: e.g. Na+ and Cl−

R [m3.atm.K−1.mol−1] gas law constant: 8.2057× 10−5

Vm [m3] volume of solvent associated with soluteT [K] temperaturec [mol of ions per m3] generic concentration

Example

Prove to yourself: 0.1 mol (∼ 1 teaspoon) of NaCl dissolved in 1 Lof water at 25◦C is 4.9 atm!

I that’s almost 500 kPaI or almost 50m of head for 5.8 g NaCl in a litre of waterI (recall: 1 atm ≈ 10 m of water height)

57

Other osmotic values

The previous equation is an approximation.

Some actual values:

Substance Osmotic pressure [atm]Pure water 0.00.1 mol NaCl in 1 L water 4.562.0 mol NaCl in 1 L water 96.2Seawater [3.5 wt% salts] 25.2

I Driving force in membrane separation is pressure difference

I ∆P = π implies we only counteract the osmotic pressure

I Reverse osmosis occurs when we increase ∆P > π

I So the net useful driving force applied: ∆P − πI Ultrafiltration ∆P was 0.1 to 1.0 MPa (10 atm) typically

I RO: typical ∆P values: 2.0 MPa to 8.0 MPa, even 10.5 MPa

58

Let’s be a little more accurate

I The solute (salt) passes through the membrane to thepermeate side

I Cp 6= 0

I There is an osmotic pressure, πperm back into the membrane.I Correct, net driving force = ∆P −∆π

I ∆P is the usual TMP we measureI ∆π = πfeed − πperm

I ∆π = Cions,feedRTfeed − Cions,permRTperm

I Even more correctly: ∆π = cions,wallRTwall − cions,permRTperm

I Draw a picture

Key point

There’s a natural limitation here: what if we try to recover toomuch solvent (high solvent flux)?

59

Widest application for RO: desalination

Some quotes:

I “McIlvaine forecasts that world RO equipment and membranesales will reach $5.6 billion (USD) in 2012, compared to $3.8billion in 2008 (actual).”

I “Depleting water supplies, coupled with increasing waterdemand, are driving the global market for desalinationtechnology, which is expected to reach $52.4 billion by 2020,up 320.3% from $12.5 billion in 2010. According to a recentreport from energy research publisher SBI Energy, membranetechnology reverse osmosis will see the largest growth,reaching $39.46 billion by 2020.”

60

Industrial applications of RO

I demineralization of industrial water before ion exchange

I not primary aim, but RO membranes retain > 300 Daltonorganics

I ultrahigh-purity waterI laboratoriesI kidney dialysisI microelectronic manufacturingI pharmaceutical manufacturing (purified water)

I tomato, citrus, and apple juice dewatering [∼ 4.5 c/L; 1995]

I dealcoholization of wine and beer to retain flavour in theretentate

I other: keep antifreeze, paint, dyes, PAH, pesticides inretentate; discharge permeate to municipal wastewater

61

Salt-water reverse osmosis example

I Larnaca, Cyprus [island state near Greece/Turkey]I Desalination plant: Build, Own, Operate, and TransferI 21.5 million m3 per yearI Seawater intake → flocculation and filtration [why?] → RO→ chemical dosing → chlorination

I Energy recovery of ∆P (see http://www.youtube.com/watch?v=M3mpJysa6zQ: novel

way of recovering pressure energy)

[ide-tech.com]

62

RO costs [Perry’s; 8ed], 1992Household ROcost:

I $ 0.015 to$0.07/L

63

Transport modelling of RO

Symbolically:

J =(permeability)(driving force)

thickness= (permeance)(driving force) =

driving force

resistance

I permeability =f (membrane properties, diffusivity, other physical properties)

I permeance: easier to calculate:I given the driving force (easy to measure)I given the flux (easy to measure)

I units are always case specific and must be self-consistent[check!]

64

Simplified RO modellingI We don’t consider “cake build-up”: we assume that solid

particles are mostly removed in an upstream separation stepI So ∆P overcomes osmotic pressure and membrane resistance1. Solvent flux

Jv = Jsolv =(∆P −∆π)

Rm,v +���* 0

Rcp,v

=Psolv

`M(∆P−∆π) = Asolv (∆P −∆π)

Rm,v = f (`M = membrane’s thickness, diffusivity of solvent in membrane)

2. Solute (salt) flux

Jsalt =(permeability)(driving force)

resistance=

Psalt

`M(Cw − Cp) = Asalt (Cw − Cp)

I Our assumption: Cwall ≈ Cbulk = bulk solute concn = CwI how would you enforce this reasonable assumption?

I Crudely assume: Cbulk ≈ Cfeed [for back-of-envelope calculations]

I Psolv = permeability of the solvent [notation: Psolv ≡ Pw]I Cp = concentration of solute in the permeateI Psalt = permeability of the salt through membrane 65

Example to try

Brackish water of 1.8wt% NaCl at 25◦C and 1000 psia [68.5 atm]is fed to a spiral wound membrane.

Conditions on the permeate side are 0.05 wt% NaCl, at sametemperature, but 50 psia [3.42 atm]

The permeance of water has been established as1.1× 10−4 kg.s−1.m−2.atm−1 [how would you get this number?]and 16× 10−8m.s−1 for salt is determined experimentally.

1. Calculate the flux of water in LMH.

2. What is the flux of salt through the membrane?

3. How do these fluxes compare?

4. Calculate the rejection coefficient, R, for salt.

66

Another example: calculating permeances

At 25 ◦C in a lab membrane with area A = 2× 10−3 m2 we feed asolution of 10 kg NaCl per m3 solution so well mixed and sorapidly that essentially it has the same strength leaving.

The permeate is measured as 0.39 kg NaCl per m3 solution at arate of 1.92× 10−8 m3.s−1 when applying a constant pressuredifference of 54.42 atm.

Calculate the permeance constants for solvent and salt (these werepreviously given, this example shows how to calculate themexperimentally), as well as the rejection coefficient.

67

Some questions to consider

1. What happens, in terms of osmosis, on a really hot day tofluid flow in a tree?

2. Is Psolv going to change if we use a different solute?

3. If we double the pressure drop, will we double the solvent flux?

4. Why did we not take osmotic pressure in account formicrofiltration and ultrafiltration?

5. In RO: what will be the expected effect of increasingoperating temperature?

68

Some old and new terminology

Recall from ultrafiltration:

I R = 1− CP

Cfeed= 1− CP

C0

I This rejection coefficient, R, also applies to reverse osmosis.

I A new term = cut = conversion = recovery = θ =QP

Q0is

between 40 and 50% typically

69